Problem 108 - Fundamentals of Physics

Question

The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h} .\) (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?

Step-by-Step Solution

Answer

(a) 7347 meters; (b) 79.7 km/h.

Step 1: Understand the Problem

The exercise involves two main parts related to circular motion. In part (a), we're asked to find the smallest radius of curvature that limits the acceleration to a certain value while maintaining the train's average speed. In part (b), we need to determine the speed that allows the train to stay within the acceleration limit given a specific curve radius.

Step 2: Identify the Formula for Centripetal Acceleration

The formula for centripetal acceleration is given by \( a = \frac{v^2}{r} \), where \( v \) is the speed, \( r \) is the radius of curvature, and \( a \) is the centripetal acceleration. The acceleration limit given is \( 0.050 \text{ g} \), which needs to be converted to \( \text{m/s}^2 \).

Step 3: Convert Units for Acceleration and Speed

Convert the acceleration limit from \( \text{g} \) to \( \text{m/s}^2 \): \( 0.050 \times 9.8 \approx 0.49 \text{ m/s}^2 \). The train's speed is given in km/h, which needs to be converted to m/s: \( 216 \text{ km/h} \times \frac{1000}{3600} \approx 60 \text{ m/s} \).

Step 4: Solve Part (a) for Radius

Use the centripetal acceleration formula \( a = \frac{v^2}{r} \). Rearrange it to solve for \( r \): \( r = \frac{v^2}{a} \). Substitute \( v = 60 \text{ m/s} \) and \( a = 0.49 \text{ m/s}^2 \) to find \( r \):\[r = \frac{(60)^2}{0.49} = \frac{3600}{0.49} \approx 7346.94 \text{ meters}\]Thus, the smallest radius is approximately 7347 meters.

Step 5: Solve Part (b) for Speed

Again using \( a = \frac{v^2}{r} \), rearrange to solve for \( v \): \( v = \sqrt{a \cdot r} \). Given the radius \( r = 1000 \text{ m} \) and \( a = 0.49 \text{ m/s}^2 \), calculate \( v \):\[v = \sqrt{0.49 \times 1000} = \sqrt{490} \approx 22.14 \text{ m/s}\]Convert the speed back to km/h: \( 22.14 \times \frac{3600}{1000} \approx 79.7 \text{ km/h} \).

Key Concepts

Circular Motion

Circular motion is a type of motion where an object moves along the circumference of a circle. When a train navigates a curved track, it undergoes circular motion. This kind of motion is characterized by a continuous change in the direction of the velocity, even if the speed is constant.
The main force at work in circular motion is centripetal force, which acts towards the center of the circle. This force depends on the object's speed and the radius of the circle. In our example, the passengers on the train experience this force as acceleration, which keeps them on the curved path instead of moving in a straight line. If the train goes too fast or the curve is too sharp, this acceleration can become uncomfortable or even unsafe for passengers.
Circular motion concepts are vital in designing safe and comfortable train routes, amusement park rides, and many other applications in daily life.

Radius of Curvature

The radius of curvature is a measure that represents the "sharpness" of a curve in a path. In simpler terms, the larger the radius, the gentler the curve is. Conversely, a smaller radius represents a tighter curve.
For trains like the TGV, maintaining a reasonable radius of curvature is crucial. This is because the centripetal acceleration, which passengers feel, is determined by both the speed of the train and the radius of the curve. The relationship between these quantities is given by the formula: \[ a = \frac{v^2}{r} \] where \(a\) is the centripetal acceleration, \(v\) is the speed, and \(r\) is the radius of curvature.
In practical terms, a high-speed train should have curves with large radii to limit the acceleration experienced by passengers. This makes them feel more comfortable and ensures their safety during the journey.

Speed Conversion

Speed conversion is an important step when solving physics problems involving motion, especially when different units are used. In physics, speeds are often required in meters per second (m/s) to be consistent with other measurements like acceleration in meters per second squared (m/s²).
For instance, the train speed given in this problem is in kilometers per hour (km/h), and it needs to be converted to meters per second. The conversion is done using the factor \(\frac{1000}{3600}\), where:

• 1000 meters = 1 kilometer
• 3600 seconds = 1 hour

Applying this conversion ensures that calculations, such as those involving the centripetal acceleration formula, are accurate and consistent.
Understanding how and why to convert units is crucial in physics, ensuring that equations work correctly and results are reliable.

Physics Problem Solving

Physics problem solving is a structured approach to finding solutions to complex questions involving natural phenomena. The process typically involves several key steps that ensure clarity and accuracy in the solution.
First, understanding the problem is crucial. Carefully analyze what is given and what needs to be found. Breaking down a complex problem into simpler parts, like identifying variables and constraints, can make the task more manageable.
Establish the relevant formulas and their meanings. In our example, recognizing that centripetal acceleration ties speed and radius with the acceleration experienced by passengers is vital.
Next, ensure all units are consistent. Convert measurements as necessary to maintain a cohesive and accurate calculation process.
Finally, solving the equations step-by-step, checking the intermediate results, and validating the final solution against the context of the problem can help mitigate errors.
Developing these problem-solving skills is not just important for physics, but for critical thinking and real-world problem-solving as well.